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Aluffi - Algebra, Chapter 0
AMS GSM 104 Algebra Errata available online: http://www.math.fsu.edu/~aluffi/algebraerrata/Errata.html "This text presents an introduction to algebra suitable for upper-level undergraduate or beginning graduate courses. While there is a very extensive oﬀering of textbooks at this level, in my experience teaching this material I have invariably felt the need for a self-contained text that would start ‘from zero’ (in the sense of not assuming that the reader has had substantial previous exposure to the subject), but impart from the very beginning a rather modern, categorically-minded viewpoint, and aim at reaching a good level of depth. Many textbooks in algebra satisfy brilliantly some, but not all of these requirements. This book is my attempt at providing a working alternative." =Chapter I : Preliminaries: Set theory and categories = =Chapter II : Groups, ﬁrst encounter= =Chapter III : Rings and modules= =Chapter IV : Groups, second encounter= =Chapter V : Irreducibility and factorization in integral domains= =Chapter VI : Linear algebra= =Chapter VII : Fields= =Chapter VIII : Linear algebra, reprise= =Chapter IX : Homological algebra= Section IX.1 (Un)necessary categorical preliminaries * After Example 1.11: The coproduct A \sqcup B is endowed with morphisms i_A: A \to A \sqcup B , i_B: B \to A \sqcup B which are easily checked to be monomorphisms (do this!): We have \pi_A \circ i_A = 1_A by definition, or in other words \pi_A is a left-inverse of i_A . Since monomorphisms are defined by cancellation, this is actually even a bit stronger than being a monomorphism (roughly the equivalent of "unit implies non-zerodivisor" in a commutative ring.) * Proposition 1.12, \pi_B is the cokernel of i_A: The proof given here is bizarrely stated. I had to work things out on my own, after which I eventually realized that my method matched the books. Don't, as suggested, stare at the diagram; the diagram is weird and unhelpful. Here's what it says: We already know that \pi_B \circ i_A = 0 , so what remains to show is that for any object C and any map \gamma : A \sqcup B \to C with \gamma \circ i_A = 0 , there exists a unique map \delta : B \to C such that \delta \circ \pi_B = \gamma . In particular, we claim that \delta = \gamma \circ i_B . To see this, consider the (unique!) map g : = (0 \sqcup (\gamma \circ i_B)) : A \sqcup B \to C ; that is, the unique map such that g \circ i_A = 0 and g \circ i_B = \gamma \circ i_B . By definition, \gamma satisfies both equations, so g = \gamma . On the other hand, we can also check that (\gamma \circ i_B \circ \pi_B) satisfies both equations, so g = \gamma \circ i_B \circ \pi_B . Therefore \gamma = \gamma \circ i_B \circ \pi_B = \delta \circ \pi_B , which is what we wanted to show. * Exercise 1.14: The word 'sheaf' should be replaced by 'presheaf' here. Section IX.2 Working in abelian categories Section IX.3 Complexes and homology, again Section IX.4 Cones and homotopies Section IX.5 The homotopic category. Complexes of projectives and injectives Section IX.6 Projective and injective resolutions, and the derived category Section IX.7 Derived functors Section IX.8 Double complexes Section IX.9 Further topics Category:Textbook Category:Algebra